Re: SUBSTRING for a regular expression
От | |
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Тема | Re: SUBSTRING for a regular expression |
Дата | |
Msg-id | 4269.192.168.123.254.1089203897.squirrel@$HOSTNAME обсуждение исходный текст |
Ответ на | Re: SUBSTRING for a regular expression ("Greg Sabino Mullane" <greg@turnstep.com>) |
Список | pgsql-general |
> SELECT > SUBSTRING(description FROM '(.*), \\d{1,3},') AS vname, > SUBSTRING(description FROM '.*, (\\d{1,3}),') AS age, > SUBSTRING(description FROM '\\d{1,3}, of (.*?),? was charged') > AS address, > SUBSTRING(description FROM ' was charged ([^ ]+)') AS dow, > SUBSTRING(description FROM ' was charged [^ ]+ with (.+)') AS charge > FROM police_log; > > - -- > Greg Sabino Mullane greg@turnstep.com > PGP Key: 0x14964AC8 200407062103 Thanks Greg, I was hopeful that this would work, since I had missed the need to double the back-slash escape character in my original work, but something still isn't right. First I got an error message that psql didn't like the "?" characters in the RE, so I eliminated them and wrote SELECT SUBSTRING(description FROM '(.*), \\d{1,3},') AS vname, SUBSTRING(description FROM '.*, (\\d{1,3}),') AS age, SUBSTRING(description FROM '\\d{1,3}, of (.*), was charged') AS address, SUBSTRING(description FROM ' was charged ([^ ]+)') AS dow, SUBSTRING(description FROM ' was charged [^ ]+ with (.+)') AS charge FROM police_log; This modified version of your suggestion gets the dow, and charge columns right, but I'm still not seeing the vname, age, and address columns -- they return null. You've gotten me part way there, and I appreciate that. Any further ideas? --Berend Tober
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